SOLUTION: Two solutions, one containing 4.5% iodine and the other containing 12% iodine, are to be mixed to produce 10 liters of a 6% iodine solution. How many liters of each are required?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Two solutions, one containing 4.5% iodine and the other containing 12% iodine, are to be mixed to produce 10 liters of a 6% iodine solution. How many liters of each are required?      Log On

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Question 501534: Two solutions, one containing 4.5% iodine and the other containing 12% iodine, are to be mixed to produce 10 liters of a 6% iodine solution. How many liters of each are required?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = liters of 4.5% solution needed
Let b = liters of 12% solution needed
given:
+.045a+ = iodine in 4.5% solution
+.12b+ = iodine in 12% solution
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(1) +a+%2B+b+=+10+
(2) +%28.045a+%2B+.12b+%29+%2F+10+=+.06+
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(2) +.045a+%2B+.12b+=+.6+
Multiply both sides of (2) by +1000+
(2) +45a+%2B+120b+=+600+
Divide both sides by +15+
(2) +3a+%2B+8b+=+40+
Multiply both sides of (1) by 3, and
subtract (1) from (2)
(2) +3a+%2B+8b+=+40+
(1) +-3a+-+3b+=+-30+
+5b+=+10+
+b+=+2+
and, since
(1) +a+%2B+b+=+10+
(1) +a+%2B+2+=+10+
(1) +a+=+8+
8 liters of 4.5% solution are needed
2 liters of 12% solution are needed
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check:
(2) +%28.045a+%2B+.12b+%29+%2F+10+=+.06+
(2) +%28.045%2A8+%2B+.12%2A2+%29+%2F+10+=+.06+
(2) +%28+.36+%2B+.24+%29+%2F+10+=+.06+
(2) +.6%2F10+=+.06+
(2) +.6+=+.6+
OK